Does there exist a geometric progression containing 27,8 and 12 as three of its term ? If it exists, then how many such progressions are possible ?

Explanation

Let, if possible, 8 be the first term of G.P and 12 and 27 be mth and nth terms of the same G.P., respectively.
`therefore 12=8r^(m-1)`
`rArr3/2=r^(m-1)`
Also, `27=8r^(n-1)`
`rArr(3/2)^(3)=r^(n-1)`
From (1) and (2), we get
`r^(n-1)=r^(3(m-1))`
`rArrn-1=3m-3`
`rArr3m=n+2`
`rArrm/1=(n+2)/3=k` (say)
`therefore` m=k,n=3k-2
By giving k different values, we get integral values of m and n.
Hence, there can be infinite number of G.P.s whose any three terms will be 8,12 and 27 (not consecutive).